Sure.. square both sides..

Algebra Level 3

Solve for x x in the following equation: x + x − x − x = 2 x x + x \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}}=2\sqrt{\dfrac{x}{x+\sqrt{x}}}

Where x x is a positive real number.


The answer is 1.00.

Sean Ty by Sean Ty , 21, Philippines

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2 solutions

x + x − x − x = 2 x x + x Multiply both sides by x + x x + x − x 2 − x = 2 x x − x = x 2 − x Squaring both sides x 2 − 2 x x + x = x 2 − x − 2 x x + x = − x Rearranging x ( 1 − x ) = 0 \begin{aligned} \sqrt{x+\sqrt{x}}-\sqrt{x-\sqrt{x}} & = 2 \sqrt{\dfrac {x}{x+\sqrt{x}}} & \small \color{#3D99F6} \text{Multiply both sides by }\sqrt{x+\sqrt x} \\ x + \sqrt x - \sqrt{x^2-x} & = 2\sqrt x \\ x - \sqrt x & = \sqrt{x^2 - x} & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 - 2x\sqrt x + x & = x^2 - x \\ - 2x \sqrt x + x & = -x & \small \color{#3D99F6} \text{Rearranging} \\ x(1-\sqrt x) & = 0 \end{aligned}

⟹ x = { 0 LHS: 2 x x + x is undefined. 1 − x = 0 ⟹ x = 1 \implies x = \begin{cases} 0 & \text{LHS: } 2\sqrt{\dfrac x{x+\sqrt x}} \text{ is undefined.} \\ 1 - \sqrt x = 0 & \implies x = \boxed 1 \end{cases}

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In the 3rd line it should be ⇒ x + x − x 2 − x = 2 x \Rightarrow x + \sqrt{x} - \sqrt{x^{2} - x} = 2\sqrt{x} ⇒ x − x = x 2 − x \Rightarrow x - \sqrt{x} = \sqrt{x^{2} - x}

Krutarth Patel - 2 years, 4 months ago

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I have changed the solution.

Chew-Seong Cheong - 2 years, 4 months ago
Ramiel To-ong
Sep 11, 2015

same analysis

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